\newproblem{lay:1_4_18}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.4.18}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $B=\begin{pmatrix}1 & 4 & 1 & 2 \\ 0 & 1 & 3 & -4 \\ 0 & 2 & 6 & 7 \\ 2 & 9 & 5 & -7\end{pmatrix}$. Can every row of $\mathbb{R}^4$ be
	written as a linear combination of the columns of $B$? Do the columns of $B$ span $\mathbb{R}^3$?
}
{
  % Solution
	Let's see if every column of $B$ has a pivot element. For doing so, we will compute a row-equivalent matrix by applying row elementary operations:
	\begin{center}
		$B\sim \begin{pmatrix}1 & 4 & 1 & 2 \\ 0 & 1 & 3 & -4 \\ 0 & 0 & 0 & 15 \\ 0 & 0 & 0 & 0\end{pmatrix}$
	\end{center}
	Not all the columns have a pivot element, for instance, column 3 has not, therefore, the columns of $B$ cannot spane $\mathbb{R}^4$. The columns of
	$B$ do not span $\mathbb{R}^3$ because they are vectors of $\mathbb{R}^4$ and not vectors of $\mathbb{R}^3$.
}
\useproblem{lay:1_4_18}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
